\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 45, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/45\hfil Multiple positive solutions]
{Multiple positive solutions for nonlinear second-order m-point
boundary-value problems with sign changing nonlinearities}

\author[F. Xu, Z. Chen,  F. Xu\hfil EJDE-2008/45\hfilneg]
{Fuyi Xu, Zhenbo Chen, Feng Xu}  

\address{Fuyi Xu \newline
 School of Mathematics and Information Science,
Shandong  University of Technology,
 Zibo, Shandong, 255049, China}
\email{zbxufuyi@163.com}

\address{Zhenbo Chen \newline
 School of Mathematics and Information Science,
Shandong  University of Technology,
 Zibo, Shandong, 255049, China}
\email{czb56@sdut.edu.cn}

\address{Feng Xu \newline
 School of Mathematics and Information Science,
Shandong  University of Technology,
 Zibo, Shandong, 255049, China}
\email{zbxf878@126.com}

\thanks{Submitted December 27, 2007. Published March 29, 2008.}
\thanks{Supported by grant 10471075 from the
 the National Natural Science Foundation of China}
\subjclass[2000]{34B15}
\keywords{m-point;  boundary-value problem; Green's function; 
\hfill\break\indent fixed point theorem in double cones}

\begin{abstract}
 In this paper, we study the nonlinear second-order m-point
 boundary value problem
 \begin{gather*}
 u''(t)+f(t,u)=0,\quad  0\leq t \leq 1, \\
 \beta u(0)-\gamma u'(0)=0,\quad
 u(1)=\sum _{i=1}^{m-2}\alpha_{i} u(\xi_{i}),
 \end{gather*}
 where the nonlinear term $f$ is allowed to change sign.
 We impose growth conditions  on $f$ which yield the existence of at
 least two  positive solutions by using  a  fixed-point theorem
 in double cones. Moreover, the associated Green's function for
 the above problem is given.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

The study of multi-point boundary value problems for linear
second-order ordinary differential equations was initiated by
Il'in and Moviseev \cite{i1,i2}.  Motivated by the study of \cite{i1,i2},
Gupta \cite{g3} studied certain three-point boundary value problems for
nonlinear ordinary differential equations. Since then,  more
general nonlinear multi-point boundary value problems have been
studied by several authors. We refer the reader to
\cite{g4,l1,m1} for
some references along this line. Multi-point boundary value
problems describe many phenomena in the applied mathematical
sciences. For example,  the vibrations of a guy wire composed of
$N$ parts with a uniform cross-section throughout but different
densities in different parts can be set up as a multi-point
boundary value problems (see \cite{m3}). Many problems in the theory of
elastic stability can be handle by the method of multi-point
boundary value problems (see \cite{h1}).

In 1997, Henderson and Wang \cite{h1} studied the existence of positive
solutions for nonlinear eigenvalue problem
\begin{gather*}
u''(t)+\lambda h(t)f(u)=0,\quad  0\leq t \leq 1, \\
u(0)=0,\quad u(1)=0,
\end{gather*}
where $f\in C([0,+\infty),[0,+\infty))$ and $h\in C([0,1],[0,+\infty))$.
The authors establish the existence of positive solutions
 under the condition that $f$ is either superlinear or sublinear.

Ma \cite{m1} investigated the  second-order three-point
  boundary value problem(BVP)
\begin{gather*}
u''(t)+a(t)f(u)=0,\quad  0 \leq t \leq 1,\\
 u(0)=0,\quad u(1)=\alpha u(\eta),
\end{gather*}
where $0 < \eta < 1$, $0<\alpha\eta<1$,
$f\in C ([0,+\infty),[0,+\infty))$, $a\in C([0,1], [0,+\infty))$. The
existence of at least one positive solution is obtained under the
condition that $f$ is either superlinear or sublinear by applying
Guo-Krasnoselskii's fixed point theorem.

Recently, Ma \cite{m2} studied the second-order m-point
 boundary-value  problem
\begin{gather*}
u''(t)+ a(t)f(u)=0,\quad 0 \leq t \leq 1,  \\
 u(0)=0,\quad  u(1)=\sum _{i=1}^{m-2}\alpha_{i} u(\xi_{i}),
\end{gather*}
where $\alpha_i\geq 0$, $i=1, 2, \dots, m-3$,
$\alpha_{m-2}>0$, $0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$,
$0<\Sigma_{i=1}^{m-2}\alpha_i\xi_i<1 $,
$f\in C ([0,+\infty),[0,+\infty))$, $a\in C([0,1], [0,+\infty))$. The
author obtained  the existence of at least one positive solution
if $f$ is either superlinear or sublinear by applying a fixed-point
theorem in cones.

All the above works were done under the assumption that the
nonlinear term is nonnegative,  applying the concavity of
solutions in the proofs. In this paper we study the
nonlinear second-order m-point boundary value problem
\begin{gather} \label{e1.1}
u''(t)+f(t,u)=0,\quad  0<t < 1, \\
\beta u(0)-\gamma u'(0)=0, \quad u(1)=\sum _{i=1}^{m-2}\alpha_{i} u(\xi_{i}),\label{e1.2}
\end{gather}
where the nonlinear term $f$ is allowed to change sign.
 Firstly we give the
associated Green's function for the above problem which makes
later discussions more precise. Then certain growth conditions are
imposed on $f$ which yield the existence of at least two  positive
solutions by using  a new  fixed-point theorem in double cones. In
this way we removed the usual restriction on $f\geq0$.

For a cone $K$ in a Banach space $X$ with norm $\|\cdot\|$ and a
constant $r>0$, let $K_r=\{x\in K: \|x\|<r\}$, $\partial
K_r=\{x\in K: \|x\|=r\}$. Suppose $\alpha: K\to \mathbb{R}^{+}$ is a
continuously increasing functional; i.e., $\alpha$ is continuous
and $\alpha(\lambda x)\leq \alpha(x)$ for $\lambda\in(0,1)$. Let
$$
K(b)=\{x\in K: \alpha(x)<b\},\partial K(b)=\{x\in K: \alpha(x)=b\}.
$$
and $K_{a}(b)=\{x\in K: a<\|x\|, \alpha(x)<b\}$. The origin in $X$
is denoted by $\theta$.

 Our main tool of this paper is the following fixed point
theorem in double cones.

\begin{theorem}[\cite{g1}]  \label{thm1.1}
 Let $X$ be a real  Banach  space with norm $\|\cdot\|$  and
 $K, K'\subset X$ two solid cones
 with $K'\subset K$. Suppose $T: K\to K$ and $T': K'\to K'$ are
 two  completely continuous operators and $\alpha: K'\to R^{+}$ is
 a continuously increasing functional satisfying
 $\alpha(x)\leq \|x\|\leq M  \alpha(x)$ for all $x\in K'$,
where $M\geq1$ is a constant. If
 there are constants $b>a>0$ such that
\begin{itemize}
\item[(C1)]  $\|Tx\|<a$, for $x\in\partial K_{a}$;

\item[(C2)] $\|T'x\|<a$, for $x\in\partial K'_{a}$ and
$\alpha(T'x)>b$ for $x\in \partial K'(b)$;

\item[(C3)] $Tx=T'x$, for $x\in K'_{a}(b)\cap\{u:T'u=u\}$.

\end{itemize}
Then $T$ has at least two fixed points $y_1$ and
$y_2$ in $K$, such that
$$
0\leq \|y_1\|<a<\|y_2\|, \quad \alpha(y_2)<b.
$$
\end{theorem}

\section{Preliminaries}

In this section, we present some lemmas  that are important to
prove our main results.

\begin{lemma} \label{lem2.1}
 Suppose that
$d=\beta(1-\sum_{i=1}^{^{m-2}}a_i\xi_i)
+\gamma(1-\sum_{i=1}^{^{m-2}}a_i)\neq
0$ and $y(t)\in C[0,1]$. Then boundary-value problem
\begin{gather}
u''(t)+y(t)=0,\quad  0<t < 1,\label{e2.1}\\
 \beta u(0)-\gamma u'(0)=0,\quad u(1)=\sum _{i=1}^{m-2}a_{i} u(\xi_{i}).
 \label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
\begin{aligned}
u(t)&=-\int_{0}^{t}(t-s)y(s)ds+\frac{\beta t+\gamma}{d}
\int_{0}^{1}(1-s)y(s)ds\\
&\quad -\frac{\beta t+\gamma}{d}
\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds
\end{aligned}\label{e2.3}
\end{equation}
\end{lemma}

\begin{proof}
 Integrating both sides of \eqref{e2.1} on $[0, t]$, we have
\begin{equation}
u'(t)=-\int_{0}^{t}y(s)ds+u'(0).\label{e2.4}
\end{equation}
Again integrating \eqref{e2.4} from 0 to $t$,  we get
\begin{equation}
u(t)=-\int_{0}^{t}(t-s)y(s)ds+u'(0)t+u(0).\label{e2.5}
\end{equation}
In particular,
\begin{gather*}
u(1)=-\int_{0}^{1}(1-s)y(s)ds+u'(0)+u(0), \\
u(\xi_i)=-\int_{0}^{\xi_i}(\xi_i-s)y(s)ds+u'(0)\xi_i+u(0).
\end{gather*}
By \eqref{e2.2} we get
$$
u'(0)=\frac{\beta}{d}\Big[\int_{0}^{1}(1-s)y(s)ds-
\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds\Big].
$$
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.2}
 Let $0<\sum_{i=1}^{m-2}a_i\xi_i<1$, $d>0$. If $y\in C[0,1]$ and
$y\geq0$, then the unique solution $u$ of
\eqref{e2.1}-\eqref{e2.2} satisfies $u(t)\geq0$.
\end{lemma}

\begin{proof}
 Since $u''(t)=-y(t)\leq0$, we  know that
the graph of $u(t)$ is concave down on $(0,1)$. So we only prove
$u(0)\geq0$, $u(1)\geq 0$.

Firstly, we shall prove $u(0)\geq0$ in the following  two
cases

\noindent{\bf Case i:}  If $0<\sum_{i=1}^{m-2}a_i\leq1$, by \eqref{e2.3} we
have
\begin{align*}
u(0)&=\frac{\gamma}{d}\Big[\int_{0}^{1}(1-s)y(s)ds-
\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds\big] \\
&\geq\frac{\gamma}{d}\Big[\int_{0}^{1}(1-s)y(s)ds-
\sum_{i=1}^{m-2}a_i\int_{0}^{1}(1-s)y(s)ds\Big] \\
&=\frac{\gamma}{d}\Big(1-\sum_{i=1}^{m-2}a_i\Big)
\int_{0}^{1}(1-s)y(s)ds\geq0.
\end{align*}

\noindent{\bf Case ii:}  If $\sum_{i=1}^{m-2}a_i>1$, by \eqref{e2.3} we have
\begin{align*}
u(0)&=\frac{\gamma}{d}\Big[\int_{0}^{1}(1-s)y(s)ds-
\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds\Big] \\
&\geq\frac{\gamma}{d}\Big[\int_{0}^{1}(1-s)y(s)ds-
\sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)y(s)ds\Big] \\
&=\frac{\gamma}{d}\int_{0}^{1}
\Big[(1-\sum_{i=1}^{m-2}a_i\xi_i)+(\sum_{i=1}^{m-2}a_i-1
)s\Big]y(s)ds\geq0.
\end{align*}
On the other hand, by \eqref{e2.3} we have
\begin{align*}
u(1)
&=-\int_{0}^{1}(1-s)y(s)ds+
\frac{\beta +\gamma}{d}\int_{0}^{1}(1-s)y(s)ds\\
&\quad -\frac{\beta +\gamma}{d}\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)y(s)ds \\
&\geq\frac{\beta}{d}\Big[\sum_{i=1}^{m-2}a_i
\int_{0}^{\xi_i}(\xi_i(1-s)-(\xi_i-s))y(s)ds
+\sum_{i=1}^{m-2}a_i\xi_i\int_{\xi_i}^{1}(1-s)y(s)ds\Big] \\
&\quad +\frac{\gamma}{d}\sum_{i=1}^{m-2}a_i
\Big[\int_{0}^{1}(1-s)y(s)ds-\int_{0}^{1}(\xi_i-s)y(s)ds \Big] \\
&=\frac{\beta}{d}\sum_{i=1}^{m-2}a_i
\Big[\int_{0}^{\xi_i}(1-\xi_i)sy(s)ds+
\xi_i\int_{\xi_i}^{1}(1-s)y(s)ds\Big] \\
&\quad +\frac{\gamma}{d}\sum_{i=1}^{m-2}a_i
\Big[\int_{0}^{1}(1-\xi_i)y(s)ds\Big]\geq0.
\end{align*}
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.3}
Let $\sum_{i=1}^{m-2}a_i\xi_i>1$, $d\neq0$.
If $y\in C[0,1]$ and $y\geq0$, then \eqref{e2.1}-\eqref{e2.2}
has no positive solution.
\end{lemma}

\begin{proof}
On the contrary, suppose that \eqref{e2.1}-\eqref{e2.2} has
a positive solution  $u$, then $u(\xi_i)>0$, $i=1,2,\dots, m-2$ and
$$
u(1)=\sum_{i=1}^{m-2}a_i u(\xi_i)=\sum_{i=1}^{m-2}a_i
\xi_i\frac{u(\xi_i)}{\xi_i}\geq\sum_{i=1}^{m-2}a_i
\xi_i\frac{u(\overline{\xi})}{\overline{\xi}}
>\frac{u(\overline{\xi})}{\overline{\xi}},
$$
where $\overline{\xi}=\min\{\xi_1, \xi_2, \dots ,\xi_{m-2}\}$
satisfies
$$
\frac{u(\overline{\xi})}{\overline{\xi}}
=\min\big\{\frac{u(\xi_1)}{\xi_1},\frac{u(\xi_2)}{\xi_2},
\dots,\frac{u(\xi_{m-2})}{\xi_{m-2}} \big\},
$$
which contradicts to the concave of $u(t)$.  The proof is
complete.
\end{proof}

\begin{lemma} \label{lem2.4}
Let $a_i\geq0$, $ i=1, \dots, m-2$,  $0<\sum_{i=1}^{m-2}a_i\xi_i<1$,
$d>0$. If  $y\in C[0,1]$ and  $y\geq0$, then the unique positive solution
$u(t)$ of \eqref{e2.1}-\eqref{e2.2} satisfies
$$
\inf_{t\in[\xi_{m-2}, 1]}u(t)\geq \sigma \|u\|,
$$
where
$$
\sigma=\min\big\{ \frac{a_{m-2}(1-\xi_{m-2})}{1-a_{m-2}\xi_{m-2}},\
a_{m-2}\xi_{m-2},\  \xi_{m-2}\big\}, \quad
\|u\|=\sup_{t\in[0,1]}|u(t)|.
$$
\end{lemma}


\begin{proof} Let
$u(\overline{t})=\max_{t\in[0,1]}u(t)=\|u\|$,  we shall discuss it
from the following two cases:

\noindent\textbf{Case 1:}  If $0<\sum_{i=1}^{m-2}a_i<1$.
Firstly,  assume  that $\overline{t}<\xi_{m-2}<1$, so that
$\min_{t\in[\xi_{m-2}, 1]}u(t)=u(1).$ By $u(1)=\sum_{i=1}^{m-2}a_i
u(\xi_i)\geq a_{m-2}u(\xi_{m-2})$ we have
\begin{align*}
u(\overline{t})&\leq u(1)+\frac{u(1)-u(\xi_{m-2})}{1-\xi_{m-2}}(0-1)\\
&=u(1)-\frac{1}{1-\xi_{m-2}}u(1)
+\frac{1}{1-\xi_{m-2}}u(\xi_{m-2}) \\
&\leq u(1)\Big( 1- \frac{1}{1-\xi_{m-2}}+\frac{1}{a_{m-2}(1-\xi_{m-2})}\Big) \\
&=u(1)\frac{1-a_{m-2}\xi_{m-2}}{a_{m-2}(1-\xi_{m-2})}.
\end{align*}
So that
\begin{equation}
\min_{t\in[\xi_{m-2},
1]}u(t)\geq\frac{a_{m-2}(1-\xi_{m-2})}{1-a_{m-2}\xi_{m-2}}\|u\|. \label{e2.6}
\end{equation}
Secondly, assume $\xi_{m-2}<\overline{t}<1$, then
$\min_{t\in[\xi_{m-2}, 1]}u(t)=u(1)$. Otherwise, we have
$\min_{t\in[\xi_{m-2}, 1]}u(t)=u(\xi_{m-2})$, then
$\overline{t}\in[\xi_{m-2},1], u(\xi_{m-2}) \geq
u(\xi_{m-1})\geq\dots\geq u(\xi_{2})\geq u(\xi_{1})$. By
$0<\sum_{i=1}^{m-2}a_i <1$ we have
$$
u(1)=\sum_{i=1}^{m-2}a_i u(\xi_i)
\leq\sum_{i=1}^{m-2}a_i u(\xi_{m-2})<u(\xi_{m-2})\leq u(1)
$$
which is a contradiction.
Since $u(t)$ is  concave,
$$
\frac{u(\xi_{m-2})}{\xi_{m-2}}\geq\frac{u(\overline{t})}{\overline{t}}\geq
u(\overline{t}).
$$
 In fact, since  $u(1)\geq a_{m-2}u(\xi_{m-2})$,
then $\frac{u(1)}{a_{m-2}\xi_{m-2}}\geq u(\overline{t}) $, which
implies
\begin{equation}
\min_{t\in[\xi_{m-2}, 1]}u(t)\geq a_{m-2}\xi_{m-2}\|u\|.\label{e2.7}
\end{equation}

\noindent\textbf{Case 2:}  If $\sum_{i=1}^{m-2}a_i>1$.
Firstly, assume $u(\xi_{m-2})\leq u(1)$, then
$\min_{t\in[\xi_{m-2}, 1]}u(t)=u(\xi_{m-2})$.  By concave of
$u(t)$ we have $\overline{t}\in[\xi_{m-2}, 1]$, which implies
$$
\frac{u(\xi_{m-2})}{\xi_{m-2}}\geq\frac{u(\overline{t})}{\overline{t}}\geq
u(\overline{t}),
$$ then
\begin{equation}
\min_{t\in[\xi_{m-2},1]}u(t)\geq\xi_{m-2}\|u\|.\label{e2.8}
\end{equation}
 Secondly, assume
$u(\xi_{m-2})> u(1)$, and so $\min_{t\in[\xi_{m-2}, 1]}u(t)=u(1)$,
and $ \overline{t}\in[\xi_{1}, 1]$. If not,
$\overline{t}\in[0,\xi_{1})$, then $u(\xi_{1})\geq \dots\geq
u(\xi_{m-2}) > u(1)$. So we have
$$
u(1)=\sum_{i=1}^{m-2}a_i u(\xi_i)>u(1)\sum_{i=1}^{m-2}a_i \geq u(1)
$$
which is a contradiction.
Since $\sum_{i=1}^{m-2}a_i>1$, there exists
 $\overline{\xi}\in\{\xi_1, \xi_2,\dots, \xi_{m-2}\}$ such that $u(\overline{\xi})\leq
 u(1)$, then $u(\xi_1)\leq u(\xi_2)\leq\dots\leq u( \xi_{m-2})\leq
 u(1)$. Since $u(t)$ is concave, we have
 $\frac{u(1)}{\xi_1}\geq\frac{u(\xi_1)}{\xi_1}
 \geq \frac{u(\overline{t})}{\overline{t}}\geq
 u(\overline{t})$, then
\begin{equation}
\min_{t\in[\xi_{m-2}, 1]}u(t)\geq\xi_{1}\|u\|.\label{e2.9}
\end{equation}
 Therefore, by \eqref{e2.6}-\eqref{e2.9} we
have
$\inf_{t\in[\xi_{m-2}, 1]}u(t)\geq \sigma \|u\|$, where
$$
\sigma=\min\big\{
\frac{a_{m-2}(1-\xi_{m-2})}{1-a_{m-2}\xi_{m-2}},\
a_{m-2}\xi_{m-2},\ \xi_{m-2}\big\}.
$$
 The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
Suppose that $d\neq 0$.  Then the  boundary value problem
\begin{gather*}
-u''(t)=0,\quad  0 <t < 1, \\
\beta u(0)-\gamma u'(0)=0,\quad
 u(1)=\sum _{i=1}^{m-2}a_{i} u(\xi_{i})
\end{gather*}
has Green's function
\begin{equation}
G(t,s)= \begin{cases}
\frac{(\beta s+\gamma)\big[(1-t)-\sum_{j=1}^{m-2}a_j(\xi_j-t)\big]}{d},\\
\text{if } 0\leq t\leq 1, \;  s\leq \xi_1, \;  s\leq t;
\\[3pt]
\frac{(\beta s+\gamma)(1-t)-\sum_{j=i}^{m-2}a_j(\xi_j-t)(\beta
s+\gamma)+\sum_{j=1}^{i-1}a_j(\beta\xi_j+\gamma)(t-s)}{d},\\
\text{if }\xi_{r-1}\leq t\leq \xi_{r},\;  2\leq r\leq m-1,\;
  \xi_{i-1}\leq s\leq \xi_{i},\; 2\leq i\leq r, s\leq t;
\\[3pt]
\frac{(\beta t+\gamma)\big[(1-s)-\sum_{j=i}^{m-2}a_j(\xi_j-s)\big]}{d},\\
\text{if } \xi_{r-1}\leq t\leq \xi_{r},\;  2\leq r\leq m-1,\;
  \xi_{i-1}\leq s\leq  \xi_{i},\; 2\leq i\leq r, t\leq s;
 \\[3pt]
\frac{(\beta t+\gamma)(1-s)}{d},\\
\text{if } 0\leq t\leq 1,\;  \xi_{m-2}\leq s\leq 1,\;
t\leq s.
\end{cases}\label{e2.10}
\end{equation}
 Here for the sake of convenience, we write $\xi_0=0, \xi_{m-1}=1$.
\end{lemma}

\begin{proof}
 If $0\leq t\leq \xi_1$, the unique solution
\eqref{e2.3} given by Lemma \ref{lem2.1} can be rewritten as
\begin{align*}
u(t)
&=\int_{0}^{t}\frac{(\beta
s+\gamma)\big[(1-t)-\sum_{j=1}^{m-2}a_j(\xi_j-t)\big]}{d}y(s)ds \\
&\quad +\int_{t}^{\xi_1}\frac{(\beta
t+\gamma)\big[(1-s)-\sum_{j=1}^{m-2}a_j(\xi_j-s)\big]}{d}y(s)ds \\
&\quad +\sum_{i=2}^{m-2}\int_{\xi_{i-1}}^{\xi_i}\frac{(\beta
t+\gamma)\big[(1-s)-\sum_{j=i}^{m-2}a_j(\xi_j-s)\big]}{d}y(s)ds
 \\
&\quad +\int_{\xi_{m-2}}^{1}\frac{(\beta
t+\gamma)(1-s)}{d}y(s)ds.
\end{align*}
Similarly, if $\xi_{r-1}\leq t\leq \xi_{r}, 2\leq r\leq m-2$, the
unique solution \eqref{e2.3} can be expressed
\begin{align*}
u(t)
&=\int_{0}^{\xi_1}
\frac{(\beta s+\gamma)\big[(1-t)-\sum_{j=1}^{m-2}a_j(\xi_j-t)\big]}{d}y(s)ds
\\
&\quad +\sum_{i=2}^{r-1}\int_{\xi_{i-1}}^{\xi_i}
\Big[(\beta s+\gamma)(1-t)-\sum_{j=i}^{m-2}a_j(\xi_j-t)(\beta s+\gamma)\\
&\quad +\sum_{j=1}^{i-1}a_j(\beta_1\xi_j+\gamma)(t-s)\Big]
\frac{y(s)}{d}ds
\\
&\quad +\int_{\xi_{r-1}}^{t}
\Big[(\beta
s+\gamma)(1-t)-\sum_{j=r}^{m-2}a_j(\xi_j-t)(\beta s+\gamma) \\
&\quad +\sum_{j=1}^{i-1}a_j(\beta\xi_j+\gamma)(t-s)\Big]
\frac{y(s)}{d} ds
\\
&\quad +\int_{t}^{\xi_{r}}
\frac{(\beta t+\gamma)\big[(1-s)-\sum_{j=r}^{m-2}a_j(\xi_j-s)\big]}{d}
y(s)ds \\
&\quad + \sum_{i=r+1}^{m-2}\int_{\xi_{i-1}}^{\xi_i}
\frac{(\beta
t+\gamma)\big[(1-s)-\sum_{j=i}^{m-2}a_j(\xi_j-s)\big]}{d} y(s)ds \\
&\quad +\int_{\xi_{m-2}}^{1}\frac{(\beta
t+\gamma)(1-s)}{d}y(s)ds.
\end{align*}
If $\xi_{m-2}\leq t\leq 1$, the unique solution \eqref{e2.3} can be given
in the form
\begin{align*}
u(t)
&=\int_{0}^{\xi_1}\frac{(\beta s+\gamma)
\big[(1-t)-\sum_{j=1}^{m-2}a_j(\xi_j-t)\big]}{d}y(s)ds \\
&\quad +\sum_{i=2}^{m-2}\int_{\xi_{i-1}}^{\xi_i}
\Big[(\beta s+\gamma)(1-t)-\sum_{j=i}^{m-2}a_j(\xi_j-t)(\beta
s+\gamma) \\
&\quad +\sum_{j=1}^{i-1}a_j(\beta\xi_j+\gamma)(t-s)\Big]
\frac{y(s)}{d_1}
ds \\
&\quad +\int_{\xi_{m-2}}^{t} \frac{(\beta
s+\gamma)(1-t)+\sum_{j=1}^{i-1}a_j(\beta\xi_j+\gamma)(t-s)}
{d}y(s)ds \\
&\quad +\int_{t}^{1}\frac{(\beta
t+\gamma)(1-s)}{d}y(s)ds.
\end{align*}
The lemma is proved.
\end{proof}

Now let $X=C[0,1]$, $K=\{u\in X:u(t)\geq0, \forall t\in[0,1]\}$,
$K'=\{u\in X: u  \hbox{ is nonnegative, concave, and
nonincreasing}\}$. Equip $X$ with the supremum norm
$\|u\|:=\sup_{t\in[0,1]}|u(t)|$. Clearly, $K,K'\subset X$
 are cones with $K'\subset K$.  For $\forall u\in K$, define
\begin{gather*}
\alpha(u)=\min_{\xi_{m-2}\leq t\leq 1}u(t),\\
(Tu)(t)=\Big(\int_{0}^{1}G(t,s)f(s, u(s))ds\Big)^{+},\quad  t\in[0,1],
\end{gather*}
where $(B)^{+}=\max\{B, 0\}$.
$$
(Au)(t)=\int_{0}^{1}G(t,s)f(s, u(s))ds, t\in[0,1],
$$
For $x\in X$, define $\theta: X\to K$ by $(\theta
u)(t)=\max\{u(t), 0\}$, then $T=\theta\circ A$. For $u\in K'$,
define
$$
(T'u)(t)=\int_{0}^{1}G(t,s)f^{+}(s, u(s))ds, \quad t\in[0,1],
$$
where $f^{+}(t, s)=\max\{f(t,s), 0\}$.

\begin{lemma} \label{lem2.6}
Let $X=C[0,1], K=\{u\in X: u\geq0\}$. Suppose
$T: X\to X$ is completely continuous. Define
$\theta: TX\to K$ by
$$
(\theta y)=\max\{y(t), \omega(t)\}, \quad  \hbox{for } y\in TX ,
$$
where $\omega\in C^{1}[0,1], \omega(t)\geq0$ is given function.
Then
$\theta \circ T: X\to K$  is also a completely continuous
operator.
\end{lemma}

\begin{proof}
 The complete continuity of $T$ implies that
$T$ is continuous and maps each bounded subset in $X$ to a
relatively compact set. Denote $\theta y$ by $\overline{y}$.

Given a function $h\in C[0,1]$, for each $\varepsilon>0$ there is
 $\delta>0$ such that
$$
\|Th-Tg\|<\varepsilon, \quad \hbox{for }
  g\in X, \|g-h\|<\delta.
$$
Since
\begin{align*}
|(\theta T h)(t)-(\theta T g)(t)|&=
|\max\{(T h)(t),\omega(t)\}-\max\{(T g)(t), \omega(t)\}| \\
&\leq |(T h)(t)-(T g)(t)|<\varepsilon,
\end{align*}
we have
$$
\|(\theta T )h-(\theta T) g\|<\varepsilon, \quad \hbox{for }
 g\in X, \|g-h\|<\delta,
$$
and so $\theta T$ is continuous.

For any arbitrary  bounded set $D\subset X$ and
for all $\varepsilon>0$, there are $y_i, i=1,2,\dots, m$ such that
$$
T D\subset\bigcup_{i=1}^{m}B(y_i,\varepsilon),
$$
where $B(y_i,\varepsilon):=\{u\in X: \|u-y_i\|<\varepsilon\}$. Then, for
for all $\overline{y}\in(\theta\circ T)D$, there is a $y\in TD$
such that $\overline{y}(t)=\max\{y(t), \omega(t)\}$. We choose
$i\in\{1, 2, \dots, m\}$ such that $\|y-y_i\|<\varepsilon$. The
fact
$$
\max_{t\in[0,1]}|\overline{y}(t)-\overline{y}_i(t)|
\leq\max_{t\in[0,1]}|y(t)-y_i(t)|,
$$
which implies $\overline{y}\in B(\overline{y}_i,\varepsilon)$.
Hence $(\theta\circ T)D$ has a finite $\varepsilon-net$ and
therefore $(\theta\circ T)D$ is
relatively compact.
\end{proof}

\section{Main results}

In this section, we present the existence of two positive
solutions for boundary value problem \eqref{e1.1}-\eqref{e1.2}
by applying a new fixed-point theorem in double cones.

Obviously, $G(t,s)\geq0$. In the following, we denote
$$
M=\max_{t\in[0,1]}\int_{0}^{1}G(t,s)ds,\quad
 n=\min_{t\in[\xi_{m-2},1]}\int_{\xi_{m-2}}^{1}G(t,s)ds.
$$
For $ t\in[\xi_{m-2},1]$, by computing we have
\begin{align*}
\int_{\xi_{m-2}}^{1}G(t,s)ds
&=\int_{\xi_{m-2}}^{t} \frac{(\beta_1
s+\gamma_1)(1-t)+\sum_{j=1}^{i-1}a_j(\beta_1\xi_j+\gamma_1)(t-s)}
{d_1}ds \\
&\quad +\int_{t}^{1}\frac{(\beta_1
t+\gamma_1)(1-s)}{d_1}ds>0.
\end{align*}
So $0<n<M$.

In the rest of the paper, we use the following
assumptions:
\begin{itemize}
\item[(H1)] $\beta\geq0$, $\gamma>0$, $\alpha_i\geq 0$,
$ i=1, 2, \dots, m-3,\alpha_{m-2}>0$,
$ 0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$,
$0<\Sigma_{i=1}^{m-2}\alpha_i\xi_i<1 $,
$d_1=\beta(1-\Sigma_{i=1}^{^{m-2}}\alpha_i\xi_i)
+\gamma(1-\Sigma_{i=1}^{^{m-2}}\alpha_i)>0$;

\item[(H2)]   $f:[0,1]\times[0,+\infty)\to R$ is
continuous and $f(t,0)\geq(\not\equiv0), t\in[0,1]$;

\item[(H3)] $h:[0,1]\to R^{+}$ is continuous.
\end{itemize}

\begin{theorem} \label{thm3.1}
Suppose that conditions
{\rm (H1)--(H3)} hold. Assume that there exist
positive numbers $a, b, d$ such that
$$
0<\big(1+\frac{\beta}{\gamma}\big)\max\Big\{1,
\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}
\Big\}d <a<\sigma b<b
$$
 such that
\begin{itemize}
\item[(H4)]  $f(t,u)\geq0$ for $(t,u)\in[0, 1]\times[d,
b]$;

\item[(H5)] $f(t,u)<\frac{a}{M}$ for
$(t, u)\in[0, 1]\times[0, a]$;

\item[(H6)]  $f(t,u)\geq\frac{\sigma
b}{n}$  for $(t, u)\in[0, 1]\times[\sigma b, b]$.

\end{itemize} Then,   \eqref{e1.1}-\eqref{e1.2} has at least two positive
solutions $u_1$ and $u_2$ such that $0\leq \|u_1\|<a<\|u_2\|$,
$\alpha(u_2)<b$
\end{theorem}

\begin{proof}
 Firstly we prove that $T$ has a fixed point
$u_1\in K$ with $0<\|u_1\|\leq a$. In fact, for all
$u\in \partial K_a$, from (H5)we have
\begin{align*}
\|Tu\|&=\max_{t\in[0,1]}\Big(\int_{0}^{1}G(t,s)f(s,
u(s))ds\Big)^{+}  \\
&\leq\max_{t\in[0,1]}\max\Big\{\int_{0}^{1}G(t,s)f(s,
u(s))ds, 0 \Big\} \\
&< \frac{a}{M}\max_{t\in[0,1]}\int_0^1G(t,s)ds
=a.
\end{align*}
The existence of
$u_1$ is proved by using (C1) Theorem \ref{thm1.1}.

Obviously, $u_1$ is a solution of \eqref{e1.1}-\eqref{e1.2} if and only if
$u_1$ is a fixed point of $A$. Next, we need to prove that $u_1$
is a solution of \eqref{e1.1}-\eqref{e1.2}. Suppose the contrary;
 i.e., there exists $t_0\in(0,1)$ such that $u_{1}(t_0)\neq(Au_1)(t_0)$.
 It must be $(Au_1)(t_0)<0=u_(t_0)$.  Let $(t_1, t_2)$ be the maximal
interval and contains $t_0$ such that $(Au_1)(t)<0$ for all
$t\in(t_1, t_2)$. Obviously, $(t_1, t_2)\neq[0,1]$ by (H2). If
$t_2<1$, then $u_1(t)\equiv0$ for $t\in[t_1, t_2]$, and
$(Au_1)(t)<0$ for $t\in(t_1, t_2)$, and  $(Au_1)(t_2)=0$. Thus,
$(Au_1)'(t_2)=0$. From (H2) we get $(Au_1)''(t)=-f(t,0)\leq 0$
for $t\in[t_1, t_2]$. So $(Au_1)'(t)\geq 0$ for $t\in[t_1, t_2]$.
We obtain $t_1=0$. On the other hand,
$\beta(Au_1)(0)-\gamma(Au_1)'(0)=0$ implies $(Au_1)'(0)\leq0$ a
contradiction. If $t_1>0$, we have $u_1(t)\equiv0$ for $t\in[t_1,
t_2]$ and $(Au_1)(t)<0$ for $t\in(t_1, t_2)$, $(Au_1)(t_1)=0$.
Thus, $(Au_1)'(t_1)\leq0$. (H2) implies
$(Au_1)''(t)=-f(t,0)\leq 0$ for $t\in[t_1, t_2]$. So $t_2=1$. From
$(Au_1)(1)=\sum _{i=1}^{m-2}\alpha_{i}(Au_1)(\xi_{i})<0$, there
exists $i_0\in\{1,2,\dots,m-2\}$ such that $(Au_1)(\xi_i)<0$ for
$i_0\leq j\leq m-2$ and $(Au_1)(\xi_j)\geq0$ for $0\leq j\leq
i_0-1$. So $\xi_j\in(t_1, 1)$ for $i_0\leq j\leq m-2$.  From the
concavity  $(Au_1)(t)$ on  $[t_1, 1]$, we have
$$
\frac{|(Au_1)(\xi_j)|}{\xi_j-1}\leq\frac{|(Au_1)(1)|}{1-t_1},
\quad\hbox{for }  i_0\leq j\leq m-2;
$$
i.e.,
$$
|(Au_1)(\xi_j)|\leq\frac{\xi_j-t_1}{1-t_1}|(Au_1)(1)|<\xi_j|(Au_1)(1)|,
\quad \hbox{for }  i_0\leq j\leq m-2.
$$
From the above inequalities, we have
$$
\sum_{j=i_0}^{m-2}\alpha_j|(Au_1)(\xi_j)|
\leq\sum_{j=i_0}^{m-2}\alpha_j\xi_j|(Au_1)(1)|<|(Au_1)(1)|.
$$
On the other hand, from $(Au_1)(1)<0$, we have
$$
|(Au_1)(1)|=|\sum_{j=1}^{m-2}\alpha_j(Au_1)(\xi_j)|
\leq\sum_{j=i_0}^{m-2}\alpha_j|(Au_1)(\xi_j)|,
$$
a contraction. Therefore $u_1$ is a solution of \eqref{e1.1}-\eqref{e1.2}
with $0<\|u_1\|<a$.

We now show that (C2) of Theorem \ref{thm1.1} is satisfied. For
$u\in\partial K'_{a}$; i.e., $\|u\|=a$. From (H5) we have
\begin{align*}
\|T'u\|&=\max_{t\in[0,1]}\int_{0}^{1}G(t,s)f^{+}(s, u(s))ds  \\
&<\frac{a}{M}\max_{t\in[0,1]}\int_0^1G(t,s)ds
=a.
\end{align*}
Whereas for $u\in \partial K'(\sigma b)$; i.e.,
$\alpha(u)=\sigma b$. For $t\in[\xi_{m-2},1]$ we have
$\sigma b\leq u(t)\leq b$. We may use condition (H6) to obtain
\begin{align*}
 \alpha(T'u)&=\min_{t\in[\xi_{m-2},1]}\int_{0}^{1}G(t,s)f^{+}(s,
u(s))ds  \\
&\geq\min_{t\in[\xi_{m-2},1]}\int_{\xi_{m-2}}^{1}G(t,s)f^{+}(s,
u(s))ds  \\&\geq \frac{\sigma
b}{n}\min_{t\in[\xi_{m-2},1]}\int_{\xi_{m-2}}^1G(t,
s)ds \\
&=\sigma b.
\end{align*}
Finally, we show that (C3) of Theorem \ref{thm1.1} is also satisfied.
Let $u\in \partial K'_{a}(\sigma b)\cap\{u: T'u=u\}$, then
$$
\|u\|>a>\big(1+\frac{\beta}{\gamma}\big)
\max\Big\{1, \frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\big\}d.
$$
We will prove
\begin{equation}
u(0)\geq\max\Big\{1, \frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\Big\}d.
\label{e3.1}
\end{equation}
Suppose this is not true, then there exists $t_0\in(0,1)$ such
that
$$
u'(t_0)>\frac{\beta}{\gamma}\max\Big\{1,
\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\Big\}d.
$$
It follows from the concavity of $u(t)$ that
$$
u'(0)\geq u'(t_0)>\frac{\beta}{\gamma}\max\Big\{1,
\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\Big\}d.
$$
So we have
\begin{align*}
0&=\beta u(0)-\gamma u'(0)\\
&<\beta\max\Big\{1,\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\Big\}d
 -\gamma\frac{\beta}{\gamma}\max\Big\{1,
\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}\big\}d
=0,
\end{align*}
which is a contradiction.

Next we claim that $u(1)\geq d$. If not, by the concavity of
$u(t)$ we have
$$
\frac{u(\xi_i)-u(1)}{1-\xi_i}\geq\frac{u(0)-u(1)}{1-0}, \quad
\hbox{for } i=1, 2, \dots, m-2;
$$
i.e., $u(0)(1-\xi_i)\leq u(\xi_i)-\xi_iu(1)$. By $u(1)=\sum
_{i=1}^{m-2}\alpha_{i} u(\xi_{i})$ we get
$$
u(0)\leq\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}
{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}u(1)
<\frac{1-\sum_{i=1}^{m-2}\alpha_i\xi_i}{\sum_{i=1}^{m-2}\alpha_i(1-\xi_i)}d,
$$
which contradicts to \eqref{e3.1} . Thus, $d\leq u(t)\leq b$ for
$t\in[0,1]$. From (H4)  we know that $f^{+}(s, u(s))=f(s,u(s))$.
This implies that $Tu=T'u$   for
$u\in \partial K'_{a}(\sigma b)\cap\{u: T'u=u\}$.
The proof is complete.
\end{proof}

\section{Applications}

 Consider the  second-order third-point boundary value problem
\begin{gather}
u''(t)+f(t,u)=0,\quad  0 <t < 1,\label{e4.1}\\
u(0)-\frac{1}{4} u'(0)=0,\quad u(1)=2 u(\frac{1}{4}),\label{e4.2}
\end{gather}
where $\beta=1$, $\gamma=\frac{1}{4}$, $m=3$, $\alpha_1=2$,
$\xi_1=\frac{1}{4}$,
$$
f(t,u)=\begin{cases}
1-16u^{2},   & 0\leq t\leq1 , 0\leq u<\frac{1}{2}, \\
 -7+8u, & 0<t<1 , \frac{1}{2}\leq u<1, \\
 1+\frac{2}{25} (u-1)^{2},  & 0\leq t\leq1 ,\; 1\leq u<6, \\
\frac{40}{11}+2(u-6)^{2}, & 0\leq t\leq1 ,\; 6\leq u<32,  \\
 2727-5(u-32)^{2}, & 0\leq t\leq1 ,\; u\geq 32.
\end{cases}
$$
Then \eqref{e4.1}-\eqref{e4.2} has at least two positive
solutions.

\begin{proof}
Let $\xi_1=\frac{1}{4}$, $d=1$, $a=6$,
$b=32$. By Lemma \ref{lem2.5} we can get
$$
\int_{0}^{1}G(t,s)ds=
-\frac{1}{2}t^{2}+\frac{7}{4}t+\frac{7}{16},\quad
\int_{1/4}^{1}G(t,s)ds=
-\frac{1}{2}t^{2}+\frac{7}{8}t+\frac{1}{2}.
$$
So,
$M=\frac{27}{16}$,
$m=\frac{11}{16}$, $\sigma=\frac{1}{4}$. It is
easy see by calculating that
\begin{gather*}
f(t,u)\geq0,\quad \hbox{for } (t,u)\in [0, 1]\times[1, 32],\\
f(t,u)\leq3,\quad \hbox{for } (t,u)\in [0, 1]\times[0, 6],\\
f(t,u)\geq\frac{128}{11},\quad \hbox{for }(t,u)\in
[\frac{1}{4}, 1]\times[8, 32].
\end{gather*}
So the conditions of Theorem \ref{thm3.1} hold. Then  \eqref{e4.1}-\eqref{e4.2}
has at  least two positive solutions.
\end{proof}

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\end{document}